Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $n = \dfrac{-4p + 40}{4p^2 - 16p - 20} \times \dfrac{-p^2 + 15p - 50}{p - 10} $
Solution: First factor out any common factors. $n = \dfrac{-4(p - 10)}{4(p^2 - 4p - 5)} \times \dfrac{-(p^2 - 15p + 50)}{p - 10} $ Then factor the quadratic expressions. $n = \dfrac {-4(p - 10)} {4(p - 5)(p + 1)} \times \dfrac {-(p - 5)(p - 10)} {p - 10} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-4(p - 10) \times -(p - 5)(p - 10) } { 4(p - 5)(p + 1) \times (p - 10)} $ $n = \dfrac {4(p - 5)(p - 10)(p - 10)} {4(p - 5)(p + 1)(p - 10)} $ Notice that $(p - 5)$ and $(p - 10)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {4\cancel{(p - 5)}(p - 10)(p - 10)} {4\cancel{(p - 5)}(p + 1)(p - 10)} $ We are dividing by $p - 5$ , so $p - 5 \neq 0$ Therefore, $p \neq 5$ $n = \dfrac {4\cancel{(p - 5)}\cancel{(p - 10)}(p - 10)} {4\cancel{(p - 5)}(p + 1)\cancel{(p - 10)}} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $n = \dfrac {4(p - 10)} {4(p + 1)} $ $ n = \dfrac{p - 10}{p + 1}; p \neq 5; p \neq 10 $